Question

The sum of the first three terms in a geometric sequence is 31, the sum of the next three terms is 3875. Find the first term and the common ratio.

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• First term of GP series = a

• Common ratio of GP series = r

Given that, the sum of the first three terms of GP is 31

$\rm \: a_1 + a_2 + a_3 = 31$

$\rm \: a + ar + {ar}^{2} = 31 - - - (1)$

Now, further given that, sum of next three terms of GP series is 3875

$\rm \: a_4 + a_5 + a_6 = 3875$

$\rm \: {ar}^{3} + {ar}^{4} + {ar}^{5} = 3875 - - - (2)$

On dividing equation (2) by equation (1), we get

$\rm \: \dfrac{ {ar}^{3} + {ar}^{4} + {ar}^{5} }{a + ar + {ar}^{2} } = \dfrac{3875}{31}$

$\rm \: \dfrac{ {ar}^{3}(1 + r + {r}^{2} )}{a(1 + r + {r}^{2})} = \dfrac{3875}{31}$

$\rm \: {r}^{3} = 125$

$\rm \: {r}^{3} = {5}^{3}$

$\rm\implies \:\boxed{ \rm{ \:r = 5 \: }}$

On substituting r = 5 in equation (1), we get

$\rm \: a + 5a + 25a = 31$

$\rm \: 31a = 31$

$\rm\implies \:\boxed{ \rm{ \:a \: = \: 1 \: }}$

Hence,

• First term of GP series = 1

• Common ratio of GP series = 5

$\rule{190pt}{2pt}$

Formula Used :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {r}^{n - 1}}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the progression.

• n is the no. of terms.

• r is the common ratio.

$\rule{190pt}{2pt}$

Additional Information :-

↝ Sum of n  terms of an geometric progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a( {r}^{n} - 1)}{r - 1} \: \sf \: provided \: that \: r \: \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of GP.

• a is the first term of the progression.

• n is the no. of terms.

• r is the common ratio.